JEE Challenger
More from System of Particles and Rotational Motion
PhyHardAdv 2016 • Paper 2

Distance of Mass Moving in Radial Slot of Rotating Disc

Comprehension Passage

A frame of reference that accelerates relative to an inertial reference frame is defined as a non-inertial frame. For example, a coordinate frame attached to a circular disc rotating about a fixed vertical axis with constant angular velocity ω\omega serves as a non-inertial frame.\n\nThe force F⃗rot\vec{F}_{\text{rot}} observed in the rotating frame and the force F⃗in\vec{F}_{\text{in}} in an inertial frame on a particle of mass mm are connected by:\nF⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗,\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega},\nwhere v⃗rot\vec{v}_{\text{rot}} represents the velocity of the particle within the rotating frame, and r⃗\vec{r} is its position vector measured from the centre of the disc.\n\nConsider a smooth slot along a diameter of a disc of radius RR that rotates counter-clockwise with a constant angular speed ω\omega around its central vertical axis. We set up a coordinate system with its origin at the disc's center, where the xx-axis lies along the slot, the yy-axis is perpendicular to it, and the zz-axis coincides with the rotation axis (so ω⃗=ωk^\vec{\omega} = \omega \hat{k}). A small object of mass mm is gently placed inside the slot at position r⃗=(R/2)i^\vec{r} = (R/2)\hat{i} at time t=0t = 0 and is restricted to move purely along the slot.

Passage Diagram 1

Determine the distance rr of the block from the center at time tt.

Options

A

R4(eωt+e−ωt)\frac{R}{4}(e^{\omega t} + e^{-\omega t})

Correct
B

R2cos⁡ωt\frac{R}{2} \cos \omega t

C

R4(e2ωt+e−2ωt)\frac{R}{4}(e^{2\omega t} + e^{-2\omega t})

D

R2cos⁡2ωt\frac{R}{2} \cos 2\omega t

Topics & Concepts

System of Particles and Rotational MotionKinematicsRelative Motion

Step-by-Step Solution

To find the distance rr of the block from the center of the disc at any time tt, we analyze the motion of the block in the non-inertial rotating frame attached to the disc.

1. Motion in the Rotating Frame

The position of the block along the radial slot (xx-axis) is given by: r⃗=xi^\vec{r} = x \hat{i}

The velocity of the block relative to the rotating frame is: v⃗rot=dxdti^=x˙i^\vec{v}_{\text{rot}} = \frac{dx}{dt} \hat{i} = \dot{x} \hat{i}

The angular velocity vector of the disc is: ω⃗=ωk^\vec{\omega} = \omega \hat{k}

The equation for the force acting on the particle in the rotating frame is: F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega}

2. Analysis of the Component Forces

  • Real Forces (F⃗in\vec{F}_{\text{in}}): Since the slot is smooth, there is no frictional force along the slot (xx-axis). The normal reaction forces act along the yy-axis (from the side walls of the slot) and the zz-axis (balancing gravity). Thus, the real force along the xx-axis is zero: (F⃗in)x=0(\vec{F}_{\text{in}})_x = 0

  • Coriolis Force: 2m(v⃗rot×ω⃗)=2m(x˙i^×ωk^)=−2mωx˙j^2m(\vec{v}_{\text{rot}} \times \vec{\omega}) = 2m(\dot{x}\hat{i} \times \omega\hat{k}) = -2m\omega\dot{x}\hat{j} This force acts entirely along the yy-axis and does not affect the radial motion along the xx-axis.

  • Centrifugal Force: ω⃗×r⃗=(ωk^)×(xi^)=ωxj^\vec{\omega} \times \vec{r} = (\omega\hat{k}) \times (x\hat{i}) = \omega x \hat{j} m(ω⃗×r⃗)×ω⃗=m(ωxj^)×(ωk^)=mω2xi^m(\vec{\omega} \times \vec{r}) \times \vec{\omega} = m(\omega x \hat{j}) \times (\omega\hat{k}) = m\omega^2 x \hat{i} This force acts outwards along the xx-axis.

3. Differential Equation of Motion

Applying Newton's second law along the xx-axis in the rotating frame: md2xdt2=mω2xm \frac{d^2x}{dt^2} = m\omega^2 x

Dividing by mm: d2xdt2−ω2x=0\frac{d^2x}{dt^2} - \omega^2 x = 0

The general solution to this homogeneous second-order linear differential equation is: x(t)=Aeωt+Be−ωtx(t) = A e^{\omega t} + B e^{-\omega t}

4. Applying Initial Conditions

  1. At t=0t = 0, the block is placed at a distance x(0)=R2x(0) = \frac{R}{2}: A+B=R2A + B = \frac{R}{2}

  2. At t=0t = 0, the block is gently released, meaning its relative velocity is zero, i.e., x˙(0)=0\dot{x}(0) = 0: x˙(t)=Aωeωt−Bωe−ωt\dot{x}(t) = A\omega e^{\omega t} - B\omega e^{-\omega t} x˙(0)=ω(A−B)=0  ⟹  A=B\dot{x}(0) = \omega(A - B) = 0 \implies A = B

Substituting A=BA = B into the first equation: 2A=R2  ⟹  A=B=R42A = \frac{R}{2} \implies A = B = \frac{R}{4}

5. Final Expression for Distance rr

Substituting the constants AA and BB back into x(t)x(t): r(t)=x(t)=R4(eωt+e−ωt)r(t) = x(t) = \frac{R}{4}\left(e^{\omega t} + e^{-\omega t}\right)

Thus, the correct option is A.

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